# Using a sequence argument to show that $\lim_{x\to \infty} \frac{x^3\sin(x)}{x^2+1}$ does not exist

Consider the following the function $$f(x)=\frac{x^3\sin(x)}{x^2+1}$$. What is the $$\lim_{x\to \infty} \frac{x^3\sin(x)}{x^2+1}$$?

My answer the limit doest not exist because $$x_n=n\pi$$ goes to $$0$$ as $$n\to\infty$$ and $$f(x_n)=0$$ for all $$n$$ so $$f(x_n)$$ goes to $$0$$ as $$n\to \infty.$$ Also, take $$y_n=(2n+1)\frac{\pi}{2}$$ goes to $$\infty$$ as $$n\to \infty$$ but $$f(y_n)=\frac{y_n ^3}{y_n^2+1} (-1)^{n}$$. So, $$\lim_{n\to\infty} f(y_n)$$ does not exist. Thus, $$\lim_{x\to \infty} \frac{x^3\sin(x)}{x^2+1}$$ does not exist.

Is that right? Is there another way to do it?

• Yes it is right!
– user1054388
Commented Jul 28, 2022 at 5:05
• Looks good. Just the $y_n$ on their own are sufficient to show that the limit does not exist. Commented Jul 28, 2022 at 5:06
• @copper.hat. What did you mean by $y_n$ sufficient ?
– Gob
Commented Jul 28, 2022 at 5:09
• Meaning you do not need the $x_n$. Commented Jul 28, 2022 at 5:10
• @copper.hat. Yes you are right it would be enough
– Gob
Commented Jul 28, 2022 at 5:11

Yes, your idea is correct. But you want the $$y_n = (4n+1)\dfrac{\pi}{2}$$ instead. Then $$f(y_n) \to \infty$$, and $$f(x_n) \to 0$$ where $$x_n = n\pi$$. So you have $$2$$ different limits and hence no limit !
• That should be sufficient, but the other side you should have an example that has the limit a number other than $0$. My example is fair enough to demonstrate that point. Commented Jul 28, 2022 at 5:23
• @WangYeFei Don’t you mean $f(y_n)\to\infty$? Commented Jul 28, 2022 at 6:32
• The exponent $3$ looks like a $2$. So, yes you are right, it goes to $\infty$. Commented Jul 28, 2022 at 6:53